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[Post New]12 Jun 2007 16:21:08 IST
    Subject: sove it..
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catch_arnnie (521)

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Olaaa!! Perrrfect answer. 81  [139 rates]
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i think there are 2 solutions..

this is how i've done --->

|[x]-2x|=3

=> [x] - 2x = 3 or [x] - 2x = -3

=> [x] = 2x + 3 or [x] = 2x - 3

since. LHS is an interger in the above equations, therefore RHS has to be an integer, & for that, x has to be an integer or x should be 1/2

but x = 1/2 does not give the solution

=> x has to be an interger

so, [x] = x

=> x = -3 or x = 3

therefore, there are 2 solutions

plz correct me if i'm wrong anywhere ....
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